| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3613 | Accepted: 1911 |
Description
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cowA has a greater skill level than cow
B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠
B), then cow A will always beat cowB.
Farmer John is trying to rank the cows by skill level. Given a list the results ofM (1 ≤
M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer,A, is the winner) of a single round of competition:
A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
5 5 4 3 4 2 3 2 1 2 2 5
Sample Output
2
Source
#include<stdio.h>
#include<string.h>
int n,m;
int path[105][105];
void floyd()
{
int i,j,k;
for(k=1;k<=n;k++)
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
if(path[i][k]!=-1&&path[k][j]!=-1)
{
if(path[i][j]==-1||path[i][j]<path[i][k]+path[k][j])
path[i][j]=path[i][k]+path[k][j];
}
}
}
int main()
{
int i,a,b,ans=0,j;
memset(path,-1,sizeof(path));
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
path[a][b]=1;
}
floyd();
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(i==j)
continue;
if(path[i][j]==-1&&path[j][i]==-1)
{
ans++;
break;
}
}
}
printf("%d/n",n-ans);
return 0;
}